Hdu 5047 Sawtooth（数学+大整数）

题目链接

Sawtooth

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 920    Accepted Submission(s): 343

Problem Description

Think about a plane:

● One straight line can divide a plane into two regions.
● Two lines can divide a plane into at most four regions.
● Three lines can divide a plane into at most seven regions.
● And so on...

Now we have some figure constructed with two parallel rays in the same direction, joined by two straight segments. It looks like a character “M”. You are given N such “M”s. What is the maximum number of regions that these “M”s can divide a plane ?

 

Input

The first line of the input is T (1 ≤ T ≤ 100000), which stands for the number of test cases you need to solve.

Each case contains one single non-negative integer, indicating number of “M”s. (0 ≤ N ≤ 10¹²)

 

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then an integer that is the maximum number of regions N the “M” figures can divide.

 

Sample Input

212

 

Sample Output

Case #1: 2Case #2: 19

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online 

题意：由两条平行的射线和两条线段构成的类似 ‘M’ 的图形，现在有n个，任意放，问最多能将平面分成多少个区域？

题解：由欧拉公式可知：对于一个几何结构的点数+面数-边数+1=空间维数

这题本来是个二维图，因为有射线，我们可以虚拟一个无穷远的点，把图形看成三维。那么问题就是求新图形的面数。

面数=边数+空间维数-点数-1；

n个‘M’相交，边数最多为： 

一条线最多被切成：4*（n-1）+1 条线段

每个‘M’4条线段，n个‘M’的总线段就是：

16*n^2-12*n

空间维度为：3

点数最多为： 

两个‘M’相交最多产生 16个点，一个‘M’自身有3个点，所有’M‘两两相交的话，再加上无穷远那个点，总点数是：

16*n*(n-1)/2+3*n+1=8*n^2-5*n+1

最后可得面数=8*n^2-7*n+1。

数据很大要用高精度，由于最后答案最大是10^24的数量级，也可以用两个long long 来拼成一个大整数。

高精度解法的代码：

//#pragma comment(linker, "/STACK:102400000,102400000")#include<stdio.h>#include<queue>#include<iostream>#include<algorithm>#include<string.h>#include<string>#include<math.h>#include<stack>typedef long long LL;using namespace std;struct BigInt{    const static int mod = 10000;    const static int DLEN = 4;    int a[600],len;    BigInt()    {        memset(a,0,sizeof(a));        len = 1;    }    BigInt(long long v)    {        memset(a,0,sizeof(a));        len = 0;        do        {            a[len++] = v%mod;            v /= mod;        }while(v);    }    BigInt(const char s[])    {        memset(a,0,sizeof(a));        int L = strlen(s);        len = L/DLEN;        if(L%DLEN)len++;        int index = 0;        for(int i = L-1;i >= 0;i -= DLEN)        {            int t = 0;            int k = i - DLEN + 1;            if(k < 0)k = 0;            for(int j = k;j <= i;j++)                t = t*10 + s[j] - '0';            a[index++] = t;        }    }    BigInt operator +(const BigInt &b)const    {        BigInt res;        res.len = max(len,b.len);        for(int i = 0;i <= res.len;i++)            res.a[i] = 0;        for(int i = 0;i < res.len;i++)        {            res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);            res.a[i+1] += res.a[i]/mod;            res.a[i] %= mod;        }        if(res.a[res.len] > 0)res.len++;        return res;    }    BigInt operator *(const BigInt &b)const    {        BigInt res;        for(int i = 0; i < len;i++)        {            int up = 0;            for(int j = 0;j < b.len;j++)            {                int temp = a[i]*b.a[j] + res.a[i+j] + up;                res.a[i+j] = temp%mod;                up = temp/mod;            }            if(up != 0)                res.a[i + b.len] = up;        }        res.len = len + b.len;        while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;        return res;    }    void output()    {        printf("%d",a[len-1]);        for(int i = len-2;i >=0 ;i--)            printf("%04d",a[i]);        printf("\n");    }}ans;int main(){    int t,cas=1;    LL n;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        ans=BigInt(8*n-7)*BigInt(n)+BigInt(1);        printf("Case #%d: ",cas++);        ans.output();    }    return 0;}

用两个long long 拼大整数的做法如下：

//#pragma comment(linker, "/STACK:102400000,102400000")#include<stdio.h>#include<queue>#include<iostream>#include<algorithm>#include<string.h>#include<string>#include<math.h>#include<stack>typedef long long LL;using namespace std;const int mod=1000000;LL n;int main(){    int t,cas=1;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        LL ltem,rtem;        ltem=(8*n-7)/mod,rtem=(8*n-7)%mod;        LL lans=0,rans;        rans=(rtem*n)%mod;        lans=ltem*n+(rtem*n)/mod;        rans=(rans+1)%mod;        lans=lans+(rans+1)/mod;        printf("Case #%d: ",cas++);        if(lans)        {            printf("%I64d%06I64d\n",lans,rans);        }        else            printf("%I64d\n",rans);    }    return 0;}